Quiz #3

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This quiz is due at the beginning of class on the 24th of November (Monday). You are on your honor not to consult others or reference materials in the completion of this quiz. This includes textbooks and notes. Please note that there are two sides to the quiz.

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1.) (15 Points) A spool of string of mass m, moment of inertia, I, and radius R, unwinds under the force of gravity. In other words, one end of the string is fixed to the ceiling and the spool of string unwinds as it falls downward.

a) What is the total torque on the system?

b) By using conservation of energy, find the velocity of the center of mass of the spool after a length h of string has unwound. Recall that

$I_{spool} = \frac{1}{2} m R^{2}$   (about the center of the cylinder)
$v = R \omega$

Solution:
We know that torque is given by several formulas. Any one of them, if used properly would be worth full credit on part a). For example:
$\vec{\tau} = \vec{r} \times \vec{F}$
$\tau = TR$
where T is the tension in the string. Another possibility is
$\vec{\tau} = I \vec{\alpha}$
$\tau = \frac{1}{2} m R^{2} \alpha$
one can then sum the forces and solve for the acceleration of the center of mass of the spool and convert that to the angular acceleration, although not necessary. The key thing to realize is that gravity does not generate a torque on the system.

In part b), conservation of energy yields
U_i + K_ri + K_ti = U_f + K_rf + K_tf
where K_r is the rotational kinetic energy and K_t is the translational kinetic energy. You need both of these terms. Without the translational term, the center of mass of the spool doesn't move and the spool just sits there and spins.
$mgh = \frac{1}{2}I\omega^{2} + \frac{1}{2}mv^{2}$
$mgh = \frac{1}{4} mR^{2}(\frac{v^{2}}{R^{2}}) + \frac{1}{2}mv^{2}$
$gh = \frac{3}{4} v^{2}$
or $v = \sqrt{\frac{4}{3}gh}$.

2.) (10 Points) A solid sphere and a solid cylinder, each of mass M and radius R, start from rest and roll without slipping down an incline. The distance that they travel in the $\hat{y}$, or vertical direction, is h. Find the velocity of the center of mass of each object at the bottom of the incline in terms of R, M, and h. Which is moving faster? §´þ

$I_{sphere} = \frac{2}{5} M R^{2}$
$I_{cylinder} = \frac{1}{2} M R^{2}$
$v = R \omega$

Solution:
From the same calculation as we did in number 1, we know that the cylinder has a velocity given by
$v = \sqrt{\frac{4}{3}gh}$
Following a very similar line of reasoning, we find that for the sphere,
$mgh = \frac{1}{5} mR^{2} \omega^{2} + \frac{1}{2} mv^{2}$
$gh = \frac{7}{10} v^{2}$
or $v = \sqrt{\frac{10}{7}gh}$
Since $\frac{10}{7} \simeq 1.43$ and $\frac{4}{3} \simeq 1.33$
v_sphere > v_cylinder.